Sum of a sequence

Sum of a sequence

I need guidance for the following question.
Using the fact that $\sum_1^{\infty}\frac{(-1)^{n+1}}{n}=\log2$,
$\sum_1^{\infty}\frac{(-1)^{n}}{n(n+1)}$ equals
$1.$ $1-2\log2$
$2.$ $1+2\log2$
$3.$ $(\log2)^2$
$4.$ $-(\log2)^2$
The given sequence gives us $1-\frac12+\frac13-\frac14+\cdots=log2$, but I
am unable to think how this would help me to solve
$-\frac12+\frac16-\frac1{12}+\frac1{20}-\cdots$
I wish somebody could help. Thanks in advance!